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I stumbled upon this number theory problem while I was solving another problem. Here is the equation: $$3^kn + 3^{k-1} + 2^m(3^{k-1} + 2h) = 2^{m+l}n$$ where $k \geq 3, h,l,m,n\in\mathbb{N}$, $n$ is odd and $n$ is not a multiple of $3$. My impression is that it does not have a solution. However, I have not progressed on the problem anymore than that. Could you please help?

Thanks.

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  • $\begingroup$ I removed the tag exponential sum because that's about the $e$ number. $\endgroup$ – quanta Apr 15 '11 at 23:03
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I think you can use infinite descent to show that the equation does not have a solution. We know that $n= 3k+1$ for $k$ an odd integer or $n =3k+2$ for $k$ an even integer. If $(k,h,l,m,n)$ is a solution, then $2^{m+1}$ divides the LHS and $n$ divides the LHS.

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  • $\begingroup$ it is a good suggestion but I do not yet see a way out since the coefficients of $n$ are not similar powers. $\endgroup$ – Chulumba Mar 16 '11 at 18:12
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    $\begingroup$ This general form of infinite descent is equivalent to mathematical induction (it is simply a contrapositive reformulation of it). So without any specific details, it is not saying much to speculate that induction might play some role. It does for almost all analogous problems - whether explicitly, or implicitly in various lemmas invoked. $\endgroup$ – Bill Dubuque Mar 16 '11 at 18:39
  • $\begingroup$ @Bill, that is a good point. $\endgroup$ – Chulumba Mar 16 '11 at 19:14
  • $\begingroup$ Thanks for the link on infinite descent. Great outline. $\endgroup$ – John Jun 13 '13 at 22:13

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